| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hi, I know this is not the maple conference, but here goes. How would I use maple to solve the following initial value problem? y'(t) = y(1-tan(t)) 1 <= t <= 2 y(0) = n = Thanks for any and all replies. scott
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1335.1 | GUESS::DERAMO | Dan D'Eramo | Mon Nov 12 1990 09:24 | 20 | |
dy -- = (1 - tan t)dt y or ln y = t + ln( cos t ) + C At t=0, ln y(0) = 0 + ln( cos 0 ) + C = ln 1 + C = C, so ln y = t + ln ( cos t ) + ln y(0) or t y(t) = y(0) e cos t I don't know how to do it in Maple. :-) Dan | |||||
| 1335.2 | Oops. | CSC32::S_JOHNSON | Lifetime Member of Aye Phelta Thi | Mon Nov 12 1990 11:53 | 8 |
The base note read:
> y'(t) = y(1-tan(t)) 1 <= t <= 2 y(0) = n =
It should be:
y'(t) = y(1-tan(t)) 1 <= t <= 2 y(1) = 0
| |||||
| 1335.3 | GUESS::DERAMO | Dan D'Eramo | Mon Nov 12 1990 14:06 | 21 | |
Did you ask in the Maple conference (MODEL::MAPLE)? Changing the initial condition to y(1) = 0 yields ln y = t + ln( cos t ) + C and so as before t C y = k e cos t k = e Now plugging in y(1) = 0 implies k = 0, so the solution is y(t) = 0 for 1 <= t <= 2. What is happening is that >> y'(t) = y(1-tan(t)) 1 <= t <= 2 y(1) = 0 says that when t=1 you have y' = 0, so y is starting at zero and staying there. Dan | |||||